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javadev merged 3 commits into javadev:main from ThanhNIT:tasks-371-372-373
Dec 16, 2021
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43 changes: 43 additions & 0 deletions src/main/java/g0301_0400/s0371_sum_of_two_integers/Solution.java
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package g0301_0400.s0371_sum_of_two_integers;

// #Medium #Top_Interview_Questions #Math #Bit_Manipulation

public class Solution {
public int getSum(int a, int b) {
int ans = 0;
int memo = 0;
int exp = 0;
int count = 0;
while (count < 32) {
int val1 = a & 1;
int val2 = b & 1;
int val = sum(val1, val2, memo);
memo = val >> 1;
val = val & 1;
a = a >> 1;
b = b >> 1;
ans = ans | (val << exp);
exp = plusOne(exp);
count = plusOne(count);
}
return ans;
}

private int sum(int val1, int val2, int val3) {
int count = 0;
if (val1 == 1) {
count = plusOne(count);
}
if (val2 == 1) {
count = plusOne(count);
}
if (val3 == 1) {
count = plusOne(count);
}
return count;
}

private int plusOne(int val) {
return (-(~val));
}
}
21 changes: 21 additions & 0 deletions src/main/java/g0301_0400/s0371_sum_of_two_integers/readme.md
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371\. Sum of Two Integers

Medium

Given two integers `a` and `b`, return _the sum of the two integers without using the operators_ `+` _and_ `-`.

**Example 1:**

**Input:** a = 1, b = 2

**Output:** 3

**Example 2:**

**Input:** a = 2, b = 3

**Output:** 5

**Constraints:**

* `-1000 <= a, b <= 1000`
70 changes: 70 additions & 0 deletions src/main/java/g0301_0400/s0372_super_pow/Solution.java
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package g0301_0400.s0372_super_pow;

// #Medium #Math #Divide_and_Conquer

public class Solution {
private static final int MOD = 1337;

public int superPow(int a, int[] b) {
int phi = phi(MOD);
int arrMod = arrMod(b, phi);
if (isGreaterOrEqual(b, phi)) {
// Cycle has started
// cycle starts at phi with length phi
return exp(a % MOD, phi + arrMod);
}
return exp(a % MOD, arrMod);
}

private int phi(int n) {
float result = n;
for (int p = 2; p * p <= n; p++) {
if (n % p > 0) {
continue;
}
while (n % p == 0) {
n /= p;
}
result *= 1.0 - 1.0 / p;
}
if (n > 1) {
// if starting n was also prime (so it was greater than sqrt(n))
result *= (1.0 - (1.0 / n));
}
return (int) result;
}

// Returns true if number in array is greater than integer named phi
private boolean isGreaterOrEqual(int[] b, int phi) {
int cur = 0;
for (int j : b) {
cur = cur * 10 + j;
if (cur >= phi) {
return true;
}
}
return false;
}

// Returns number in array mod phi
private int arrMod(int[] b, int phi) {
int res = 0;
for (int j : b) {
res = (res * 10 + j) % phi;
}
return res;
}

// Binary exponentiation
private int exp(int a, int b) {
int y = 1;
while (b > 0) {
if (b % 2 == 1) {
y = (y * a) % MOD;
}
a = (a * a) % MOD;
b /= 2;
}
return y;
}
}
36 changes: 36 additions & 0 deletions src/main/java/g0301_0400/s0372_super_pow/readme.md
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372\. Super Pow

Medium

Your task is to calculate <code>a<sup>b</sup></code> mod `1337` where `a` is a positive integer and `b` is an extremely large positive integer given in the form of an array.

**Example 1:**

**Input:** a = 2, b = \[3\]

**Output:** 8

**Example 2:**

**Input:** a = 2, b = \[1,0\]

**Output:** 1024

**Example 3:**

**Input:** a = 1, b = \[4,3,3,8,5,2\]

**Output:** 1

**Example 4:**

**Input:** a = 2147483647, b = \[2,0,0\]

**Output:** 1198

**Constraints:**

* <code>1 <= a <= 2<sup>31</sup> - 1</code>
* `1 <= b.length <= 2000`
* `0 <= b[i] <= 9`
* `b` doesn't contain leading zeros.
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package g0301_0400.s0373_find_k_pairs_with_smallest_sums;

// #Medium #Array #Heap_Priority_Queue

import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;

public class Solution {
private static class Node {
long sum;
List<Integer> al;
int index;

Node(int index, int num1, int num2) {
this.sum = (long) num1 + (long) num2;
this.al = new ArrayList<>();
this.al.add(num1);
this.al.add(num2);
this.index = index;
}
}

public List<List<Integer>> ksmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<Node> queue = new PriorityQueue<>((a, b) -> a.sum < b.sum ? -1 : 1);
List<List<Integer>> res = new ArrayList<>();
for (int i = 0; i < nums1.length && i < k; i++) {
queue.add(new Node(0, nums1[i], nums2[0]));
}
for (int i = 1; i <= k && !queue.isEmpty(); i++) {
Node cur = queue.poll();
res.add(cur.al);
int next = cur.index;
int lastNum1 = cur.al.get(0);
if (next + 1 < nums2.length) {
queue.add(new Node(next + 1, lastNum1, nums2[next + 1]));
}
}
return res;
}
}
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373\. Find K Pairs with Smallest Sums

Medium

You are given two integer arrays `nums1` and `nums2` sorted in **ascending order** and an integer `k`.

Define a pair `(u, v)` which consists of one element from the first array and one element from the second array.

Return _the_ `k` _pairs_ <code>(u<sub>1</sub>, v<sub>1</sub>), (u<sub>2</sub>, v<sub>2</sub>), ..., (u<sub>k</sub>, v<sub>k</sub>)</code> _with the smallest sums_.

**Example 1:**

**Input:** nums1 = \[1,7,11\], nums2 = \[2,4,6\], k = 3

**Output:** \[\[1,2\],\[1,4\],\[1,6\]\]

**Explanation:** The first 3 pairs are returned from the sequence: \[1,2\],\[1,4\],\[1,6\],\[7,2\],\[7,4\],\[11,2\],\[7,6\],\[11,4\],\[11,6\]

**Example 2:**

**Input:** nums1 = \[1,1,2\], nums2 = \[1,2,3\], k = 2

**Output:** \[\[1,1\],\[1,1\]\]

**Explanation:** The first 2 pairs are returned from the sequence: \[1,1\],\[1,1\],\[1,2\],\[2,1\],\[1,2\],\[2,2\],\[1,3\],\[1,3\],\[2,3\]

**Example 3:**

**Input:** nums1 = \[1,2\], nums2 = \[3\], k = 3

**Output:** \[\[1,3\],\[2,3\]\]

**Explanation:** All possible pairs are returned from the sequence: \[1,3\],\[2,3\]

**Constraints:**

* <code>1 <= nums1.length, nums2.length <= 10<sup>5</sup></code>
* <code>-10<sup>9</sup> <= nums1[i], nums2[i] <= 10<sup>9</sup></code>
* `nums1` and `nums2` both are sorted in **ascending order**.
* `1 <= k <= 1000`
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package g0301_0400.s0371_sum_of_two_integers;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void getSum() {
assertThat(new Solution().getSum(1, 2), equalTo(3));
}

@Test
void getSum2() {
assertThat(new Solution().getSum(2, 3), equalTo(5));
}
}
28 changes: 28 additions & 0 deletions src/test/java/g0301_0400/s0372_super_pow/SolutionTest.java
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package g0301_0400.s0372_super_pow;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void superPow() {
assertThat(new Solution().superPow(2, new int[] {3}), equalTo(8));
}

@Test
void superPow2() {
assertThat(new Solution().superPow(2, new int[] {1, 0}), equalTo(1024));
}

@Test
void superPow3() {
assertThat(new Solution().superPow(1, new int[] {4, 3, 3, 8, 5, 2}), equalTo(1));
}

@Test
void superPow4() {
assertThat(new Solution().superPow(2147483647, new int[] {2, 0, 0}), equalTo(1198));
}
}
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package g0301_0400.s0373_find_k_pairs_with_smallest_sums;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import java.util.ArrayList;
import java.util.Arrays;
import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void ksmallestPairs() {
assertThat(
new Solution().ksmallestPairs(new int[] {1, 7, 11}, new int[] {2, 4, 6}, 3),
equalTo(
new ArrayList<>(
Arrays.asList(
Arrays.asList(1, 2),
Arrays.asList(1, 4),
Arrays.asList(1, 6)))));
}

@Test
void ksmallestPairs2() {
assertThat(
new Solution().ksmallestPairs(new int[] {1, 1, 2}, new int[] {1, 2, 3}, 2),
equalTo(new ArrayList<>(Arrays.asList(Arrays.asList(1, 1), Arrays.asList(1, 1)))));
}

@Test
void ksmallestPairs3() {
assertThat(
new Solution().ksmallestPairs(new int[] {1, 2}, new int[] {3}, 3),
equalTo(new ArrayList<>(Arrays.asList(Arrays.asList(1, 3), Arrays.asList(2, 3)))));
}
}

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